1. Feb 6, 2005

### the_d

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A jogger runs in a straight line with an average
velocity of 4:7 m/s for 5.1 min, and then with
an average velocity of 4.1 m/s for 3 min.
What is her total displacement? Answer in
units of meters

2. Feb 6, 2005

### christinono

use the formula:
$$d = V_{average}t$$

3. Feb 6, 2005

### the_d

and then I subtract the two to find the final displacement

4. Feb 6, 2005

### dextercioby

No,you have to add the 2 distances.

Daniel.

5. Feb 6, 2005

### christinono

Well, it all depends what the direction is. If he runs in the same direction both times, you have to add the 2 displacements to find the total displacement.

6. Feb 6, 2005

thanks dude

7. Feb 6, 2005

### dextercioby

Even if the runs in the opposite direction on the second time interval,the TOTAL distance would still be gotten by adding the 2.

Daniel.

8. Feb 6, 2005

### christinono

Aren't the words "distance travelled" and "displacement" different. I though displacement meant distance and direction from the original starting point.

9. Feb 6, 2005

### the_d

for the same question, to get the avg. velocity during the time what should i do? i know it is not just to add the 2 velocities given and divide by 2

10. Feb 6, 2005

### christinono

Average velocity is total distance divided by total time.

11. Feb 6, 2005

### dextercioby

I think that u've given yourself the answer. Besides,since in the problem we're not given the directions (so that we would think it as a vector projection problem),it's natural to assume the simpler version,which is what the teacher would want,right...? :tongue2:

Daniel.

12. Feb 6, 2005

### the_d

new question

A 500-kilogram sports car accelerates uni-
formly from rest, reaching a speed of 30 me-
ters per second in 6 seconds.
During the 6 seconds, the car has traveled
a distance of how many meters??

i got 180m is that correct?
i used : distance = avg. velocity * change in time
= (30m.s) * 6
= 180m

Last edited: Feb 6, 2005
13. Feb 6, 2005

### christinono

ha ha, very funny :tongue:

14. Feb 6, 2005

### dextercioby

Daniel.

15. Feb 6, 2005

### christinono

Careful... 30m/s is not the average velocity, but rather the final velocity. Remember, average velocity is total displacement divided by the total time, or in this case (if a is constant),
$$\frac{V_i + V_f}{2} = \frac{0+30m/s}{2}$$

16. Feb 6, 2005

### the_d

where did u get the 2 from???

17. Feb 6, 2005

### christinono

Isn't the average the sum of the 2 velocities divided by 2? In this case, the displacement would be: $$d = V_{ave}t$$

18. Feb 6, 2005

### dextercioby

Kay,guys,let's leave side talks and refer to the initial problem ...

Daniel.

19. Feb 6, 2005

### the_d

your saying I take 30m/s and divide it by 2?? that is not right either.

20. Feb 6, 2005

### christinono

I's not "side talking". I'm referring to the "new question". Here, average velocity IS involved in finding the distance.

Anyways, gotta go. I hope I didn't irritate you too much Daniel